ĐKXĐ: \(x\ne\left\{-\frac{1}{2};2\right\}\)
\(A=\frac{4x^2\left(x-2\right)+3\left(x-2\right)}{\left(x-2\right)\left(2x+1\right)}=\frac{\left(4x^2+3\right)\left(x-2\right)}{\left(x-2\right)\left(2x+1\right)}=\frac{4x^2+3}{2x+1}\)
b/ \(A=\frac{4x^2-1+4}{2x+1}=2x-1+\frac{4}{2x+1}\)
Để A nguyên \(\Rightarrow\frac{4}{2x+1}\) nguyên \(\Rightarrow2x+1=Ư\left(4\right)\)
Mà \(2x+1\) lẻ \(\Rightarrow\left[{}\begin{matrix}2x+1=-1\\2x+1=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)